Using all
with generator expression, you don't need to count, compare length:
>>> [i for i in x if all(j.isdigit() or j in string.punctuation for j in i)]
['12,523', '3.46']
BTW, above and OP's code will include strings that contains only punctuations.
>>> x = [',,,', '...', '123', 'not number']
>>> [i for i in x if all(j.isdigit() or j in string.punctuation for j in i)]
[',,,', '...', '123']
To handle that, add more condition:
>>> [i for i in x if all(j.isdigit() or j in string.punctuation for j in i) and any(j.isdigit() for j in i)]
['123']
You can make it a little bit faster by storing the result of string.punctuation in a set.
>>> puncs = set(string.punctuation)
>>> [i for i in x if all(j.isdigit() or j in puncs for j in i) and any(j.isdigit() for j in i)]
['123']